Motion in A Straight Line Test

Result

Motion in A Straight Line Test - 63

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

Using the table given below where the values of velocity at the end of t seconds for a body under linear motion are given
$$V (m\:s^{-1})$$
0

6
12
24
30
36
42
$$t (s)$$
0

2
4
8
10
12
14
What can be concluded about the motion of the body?

A
It moves with uniform speed.

B
It moves with uniform motion

C
It moves with uniform velocity

D
It moves with uniform acceleration

Solution
Question 2
1 / -0

A person sitting in the rear end of the compartment throws a ball towards the front end.The ball follows a parabolic path.The train is moving with velocity of $$20\ m/s$$.A person standing outside on the ground also observes the ball.How will the maximum heights $$y_m$$ attained and the ranges $$R$$ seen by the thrower and the outside observer compare with each other?

A
same $$y_m$$ different $$R$$

B
same $$y_m$$ and $$R$$

C
differently $$y_m$$ same $$R$$

D
differently $$y_m$$ and $$R$$

Solution

The motion of train will affect only the horizontal component of the velocity of the ball.Since , vertical component is same for both observers, the $$y_m$$ will be same ,but range R will be different.
Question 3
1 / -0

Two identical balls are shot upward one after another at an interval of $$2\ s$$ along the same vertical line with same initial velocity of $$40\ ms^{-1}$$ The height at which the balls collide is

A
$$50\ m$$

B
$$75\ m$$

C
$$100\ m$$

D
$$125\ m$$

Solution

time taken for ball to rest is
$$ v= u +at$$
$$ 0 = 40 -gt$$
$$t = 4 s$$
therefore ball 1 is at rest at height $$S = 40 \times 4 -\dfrac{1}{2}\times 10 \times 4^2 = 80 m$$
velocity of ball 2 after 2s as it was thrown after the delay of 2s
$$v = u+at$$
$$v = 40 - 10 \times 2 = 20 m/s$$
$$S_2 = 40 \times 2 - \dfrac{1}{2} \times 10 \times 2^2 = 60 m$$
now ball 1 will drop with initial velocity of 0 and acceleration g and ball 2 with initial velocity of 20 m/s and deceleration of g and the separation is 20m

$$S_1 = 0 + \dfrac{1}{2}gt^2$$
$$S_2 = 20t - \dfrac{1}{2}g t^2$$

$$S_1+S_2 = 20$$
$$\dfrac{1}{2}gt^2 + 20t - \dfrac{1}{2}gt^2 = 20$$
$$t = 1s$$
$$S_1 = 5m$$
$$S_2 = 15m $$
hence ball 2 will be at 75m when they collide.
Question 4
1 / -0

A ball is thrown vertically upwards from the ground and a student gazing out of the window sees it moving upward past him at $$\displaystyle 10 \ ms^{-1}.$$ The window is at $$15$$ m above the ground level. The velocity of ball $$3\ s$$ after it was projected from the ground is (take $$ \ g= 10 \ ms^{-2} $$)

A
$$10 \ ms^{-1},$$ up

B
$$\displaystyle 20\ ms^{-1},$$ up

C
$$\displaystyle 20\ ms^{-1},$$ down

D
$$\displaystyle 10\ ms^{-1},$$ down

Solution

Let the ball be thrown with initial velocity of 'u'.
After travelling 15 m, its velocity is 10 m/s.
Thus $${v}^{2}-{u}^{2}=-2gh$$ (acceleration acts downwards, thus -g)
or, $${10}^{2}-{u}^{2}=-300$$
giving $$u=20m/s$$
After 3 sec, $$v=u-gt = 20-30 = 10 m/s $$ downwards
Question 5
1 / -0

A lift start from rest. Its acceleration is plotted against time. When it comes to rest its height above its starting point is

A
$$20\ m$$

B
$$64\ m$$

C
$$32\ m$$

D
$$36\ m$$

Solution

From 0- 4 s the motion is with constant acceleration, the distance can be calculated from kinematics equation
$$S_{0-4}=0+\frac{1}{2}2\times 4^2= 16\; m$$
From 4- 8 s the motion is with constant velocity $$v=u+at=0+2\times 4=8\; m/s$$
$$S_{4-8}=v\times t\; =8*4= 32 \; m$$
From 8- 12 s the motion is with constant acceleration, the distance can be calculated from kinematics equation
$$S_{8-12}=8\times 4-\frac{1}{2}2\times 4^2= 16\; m$$
total distance covered$$ = 16+ 32+16 m= 64 m$$
Question 6
1 / -0

A particle is projected vertically upwards and reaches the maximum height $$H$$ at a time $$t=T$$. The height of the particle at any time $$t (< T)$$ will be

A
$$\displaystyle g\left ( t-T \right )^{2}$$

B
$$\displaystyle H-g\left ( t-T \right )^{2}$$

C
$$\displaystyle \frac{1}{2}g\left ( t-T \right )^{2}$$

D
$$\displaystyle H-\frac{1}{2}g\left ( T-t \right )^{2}$$

Solution

At maximum height$$,\ v=0$$
Thus, from equations of motion:
$$0=u-gT$$ or $$u=gT$$
$$H=uT-\dfrac{1}{2}gT^2=gT^2-\dfrac{1}{2}gT^2$$
or, $$H=0.5gT^2$$
At any time$$,\ t,\ h=ut-0.5gt^2=gTt-0.5gt^2$$
Add and subtract $$0.5gT^2$$, we get,
$$h=(gtT-0.5gt^2-0.5gT^2)+0.5gT^2$$
Put $$gT^2=H$$, we get
$$h=-0.5g(T-t)^2+H=H-\dfrac{1}{2}g(T-t)^2$$
Question 7
1 / -0

A car is travelling on a straight road. The maximum velocity the car can attains is $$\displaystyle 24 ms^{-1}.$$ The maximum acceleration and deceleration it can attain are $$\displaystyle 1 ms^{-2}$$ and $$\displaystyle 4ms^{-2}$$ respectively. The shortest time the car takes from rest to rest in a distance of 500 m is,

A
22.4 s

B
30 s

C
11.2 s

D
35.8 s

Solution

$$Displacement\quad if\quad the\quad car\quad reaches\quad max\quad speed\quad with\quad max\quad acceleration\\ { s }_{ 1 }=\dfrac { { v }^{ 2 } }{ 2a } =\dfrac { { 24 }^{ 2 } }{ 2 } =288m\\ Displacement\quad when\quad the\quad car\quad reaches\quad from\quad max\quad speed\quad to\quad zero\quad \\ { s }_{ 2 }\quad =\quad \dfrac { { u }^{ 2 } }{ 2a } =\dfrac { { 24 }^{ 2 } }{ 2\times 4 } =72\\ { s }_{ 1 }+{ s }_{ 2 }=288+72=360m\\ remaining\quad distance\quad is\quad travelled\quad at\quad maximum\quad speed,\quad for\quad which\quad the\quad \\ time\quad taken\quad is\quad \dfrac { 500-360 }{ 24 } =5.833s\\ time\quad during\quad { s }_{ 1 }=\dfrac { 24 }{ 1 } =24s\\ time\quad during\quad { s }_{ 2 }=\dfrac { 24 }{ 4 } =6s\\ total\quad time\quad is\quad 35.833s$$
Question 8
1 / -0

The velocity of a particle at time $$t=0$$ is $$2m/s$$. A constant acceleration of $$2m/s^{2}$$ acts on the particle for $$2$$ seconds at an angle of $$60$$$$^{\circ}$$ with its initial velocity. The magnitude of velocity and displacement of particle at the end of $$t=2s$$ respectively are:

A
$$2$$$$\sqrt{7}$$ m/s, $$4$$$$\sqrt{3}$$ m

B
$$2$$$$\sqrt{3}$$ m/s, $$4$$$$\sqrt{7}$$ m

C
$$4$$$$\sqrt{7}$$ m/s, $$2$$$$\sqrt{3}$$ m

D
$$4$$$$\sqrt{3}$$ m/s, $$2$$$$\sqrt{7}$$ m

Solution

initial velocity $$u = 2 m/s$$
acceleration $$a$$ = $$2 m/s$$ at $$60$$$$^{\circ}$$
component of acceleration in direction of velocity and perpendicular to initiail velocity according to figure

$${ a }_{ x }=acos{ 60 }^{ 0 }\quad =\quad 2\times \dfrac { 1 }{ 2 } =\quad 1\\ { a }_{ y }=asin{ 60 }^{ 0 }\quad =\quad 2\times \dfrac { \sqrt { 3 } }{ 2 } =\sqrt { 3 } $$

$${ s }_{ x }=ut+\dfrac { 1 }{ 2 } { a }_{ x }{ t }^{ 2 }=\quad 2\times 2+\dfrac { 1 }{ 2 } \times 1{ (2) }^{ 2 }=6m\quad $$

$${ s }_{ y }=ut+\dfrac { 1 }{ 2 } { a }_{ x }{ t }^{ 2 }=\quad 0+\dfrac { 1 }{ 2 } \times \sqrt { 3 } { (2) }^{ 2 }=2\sqrt { 3 } m$$

net displacement = $$S=\sqrt { { { s }_{ x } }^{ 2 }+{ { s }_{ y } }^{ 2 } } =\quad \sqrt { 36+12 } =4\sqrt { 3 } m$$
again for velocity in X any Y direction

$${ v }_{ x }=u+{ a }_{ x }t=2+1\times 2=4m/s\\ { v }_{ y }=u+{ a }_{ y }t=0+\sqrt { 3 } \times 2=2\sqrt { 3 } m/s$$

net velocity = $$V=\sqrt { { { v }_{ x } }^{ 2 }+{ { v }_{ y } }^{ 2 } } =\quad \sqrt { 16+12 } =2\sqrt { 7 } m/s$$
thus .

option $$A$$ is correct
Question 9
1 / -0

A car moving along a straight highway with a speed of $$126 kmph$$ is brought to a stop within a distance of $$200 m$$. What is the acceleration of the car and how long does it take for the car to stop?

A
$$-3.06 ms^{-2}, 11.43 s$$

B
$$-6.03 ms^{-2}, 11.43 s$$

C
$$-3.06 ms^{-2}, 10.34 s$$

D
$$-6.03 ms^{-2}, 10.34 s$$

Solution

Initial velocity $$u = 126 km/hr$$ or $$35m/s$$
Final velocity $$=0 m/s$$ (stopped)
Distance $$s = 200 m$$
Use newton's 3rd law of motion
$$ v^{2} = u^{2} +2as$$ ; a is acceleration or -a if it is deceleration
$$0 = 35^{2} +2 \times a \times 200$$
$$400a = -1225$$
$$ a = -3.0625 m/s^{2}$$ ---------> This is retardation of the CAR to find time

Use first law of motion
$$ v = u + at$$; first law of motion,
$$ 0 = 35 - 3.0625t$$
$$ t = 35/3.0625 =11.4285 sec$$
Question 10
1 / -0

Net force acting on a particle of mass 2 kg is 10 N in north direction. At t = 0, particle was moving eastwards with 10 m/s. Find displacement and velocity of particle after 2 s.

A
$$\displaystyle 10\sqrt{5}$$ m at $$\displaystyle \cot^{-1}(2)$$ from east towards north, $$\displaystyle 10\sqrt{2}ms^{-1}\:at\:45^{\circ}$$ from east towards north.

B
$$\displaystyle 10\sqrt{2}$$ m at $$\displaystyle \cot^{-1}(2)$$ from east towards north, $$\displaystyle 10\sqrt{2}ms^{-1}\:at\:45^{\circ}$$ from east towards north.

C
$$\displaystyle 10\sqrt{5}$$ m at $$\displaystyle \cot^{-1}(2)$$ from east towards north, $$\displaystyle 10\sqrt{5}ms^{-1}\:at\:45^{\circ}$$ from east towards north.

D
$$\displaystyle 10\sqrt{2}$$ m at $$\displaystyle \cot^{-1}(2)$$ from east towards north, $$\displaystyle 10\sqrt{5}ms^{-1}\:at\:45^{\circ}$$ from east towards north.

Solution

$$a=5m/{ s }^{ 2 }along\quad north,\quad thus,\quad velocity\quad along\quad east\quad won't\quad change\\ t=2s\\ u\quad =\quad 10m/s\quad along\quad east\\ displacement\quad along\quad north\quad =\quad \frac { 1 }{ 2 } a{ t }^{ 2 }=10m\\ displacement\quad along\quad east=ut=20m\\ total\quad displacement\quad =\quad \sqrt { { 10 }^{ 2 }+{ 20 }^{ 2 } } =10\sqrt { 5 } m\\ velocity\quad along\quad north\quad =\quad at=10m/s\\ final\quad velocity\quad =\quad 10\hat { i } +10\hat { j } $$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

$$V (m\:s^{-1})$$	0		6	12	24	30	36	42
$$t (s)$$	0		2	4	8	10	12	14

Motion in A Straight Line Test - 63

Result

Motion in A Straight Line Test - 63

Score

-

Rank

-

SHARING IS CARING

Question 1

It moves with uniform speed.

It moves with uniform motion

It moves with uniform velocity

It moves with uniform acceleration

Solution

Question 2

same $$y_m$$ different $$R$$

same $$y_m$$ and $$R$$

differently $$y_m$$ same $$R$$

differently $$y_m$$ and $$R$$

Solution

Question 3

$$50\ m$$

$$75\ m$$

$$100\ m$$

$$125\ m$$

Solution

Question 4

$$10 \ ms^{-1},$$ up

$$\displaystyle 20\ ms^{-1},$$ up

$$\displaystyle 20\ ms^{-1},$$ down

$$\displaystyle 10\ ms^{-1},$$ down

Solution

Question 5

$$20\ m$$

$$64\ m$$

$$32\ m$$

$$36\ m$$

Solution

Question 6

$$\displaystyle g\left ( t-T \right )^{2}$$

$$\displaystyle H-g\left ( t-T \right )^{2}$$

$$\displaystyle \frac{1}{2}g\left ( t-T \right )^{2}$$

$$\displaystyle H-\frac{1}{2}g\left ( T-t \right )^{2}$$

Solution

Question 7

22.4 s

30 s

11.2 s

35.8 s

Solution

Question 8

$$2$$$$\sqrt{7}$$ m/s, $$4$$$$\sqrt{3}$$ m

$$2$$$$\sqrt{3}$$ m/s, $$4$$$$\sqrt{7}$$ m

$$4$$$$\sqrt{7}$$ m/s, $$2$$$$\sqrt{3}$$ m

$$4$$$$\sqrt{3}$$ m/s, $$2$$$$\sqrt{7}$$ m

Solution

Question 9

$$-3.06 ms^{-2}, 11.43 s$$

$$-6.03 ms^{-2}, 11.43 s$$

$$-3.06 ms^{-2}, 10.34 s$$

$$-6.03 ms^{-2}, 10.34 s$$

Solution

Question 10

$$\displaystyle 10\sqrt{5}$$ m at $$\displaystyle \cot^{-1}(2)$$ from east towards north, $$\displaystyle 10\sqrt{2}ms^{-1}\:at\:45^{\circ}$$ from east towards north.

$$\displaystyle 10\sqrt{2}$$ m at $$\displaystyle \cot^{-1}(2)$$ from east towards north, $$\displaystyle 10\sqrt{2}ms^{-1}\:at\:45^{\circ}$$ from east towards north.

$$\displaystyle 10\sqrt{5}$$ m at $$\displaystyle \cot^{-1}(2)$$ from east towards north, $$\displaystyle 10\sqrt{5}ms^{-1}\:at\:45^{\circ}$$ from east towards north.

$$\displaystyle 10\sqrt{2}$$ m at $$\displaystyle \cot^{-1}(2)$$ from east towards north, $$\displaystyle 10\sqrt{5}ms^{-1}\:at\:45^{\circ}$$ from east towards north.

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.